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klasskru [66]
3 years ago
12

Simplify (3x2y)(5x3y−1)

Mathematics
1 answer:
ludmilkaskok [199]3 years ago
3 0

Answer:

6xy(15xy−1)

Step-by-step explanation:

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PLEASE HELO ASAP WILL GIVE BRAINLEAST TO RIGHT ANSWER
marishachu [46]

Answer:

1/4

Step-by-step explanation:

Favorable Outcomes/ All Possible Outcomes= 2/8

2/8= 1/4

The answer is 1/4.

5 0
4 years ago
Which of these situations is represented by 24 divided by 6
sveticcg [70]
We need to see the answer choices.
3 0
3 years ago
Susan collected 89 stickers for 28 weeks. How many stickers did she collect in total round off your answer to the nearest hundre
Temka [501]

Answer:

2500 ( nearest hundred)

Step-by-step explanation:

Given that :

Number of stickers collected per week = 89

Number of weeks for which stickers were being collected = 28

To obtain the total number of stickers collected, we take the product of the number of weeks for which the stickers were beug vcilled and the number of stickers collected per week

The number of stickers collected :

89 * 28 = 2492

2492 = 2500 ( nearest hundred)

7 0
3 years ago
Melody wants to buy a present for her friend for her birthday. She wants to purchase a CD that costs $18. She thought she had $2
miv72 [106K]

Answer: 20%

Step-by-step explanation: $20-$16=$4. 4/20=1/5 or 20%.

6 0
3 years ago
Read 2 more answers
The weights of broilers (commercially raised chickens) are approximately normally distributed with mean 1387 grams and standard
yuradex [85]

Answer:

Probability that a randomly selected broiler weighs more than 1454 g is 0.3372 or 34% (approx.)

Step-by-step explanation:

Given:

Weights of Broilers are normally distributed.

Mean = 1387 g

Standard Deviation = 161 g

To find: Probability that a randomly selected broiler weighs more than 1454 g.

we have ,

Mean,\,\mu=1387

Standard\,deviation,\,\sigma=161

X = 1454

We use z-score to find this probability.

we know that

z=\frac{X-\mu}{\sigma}

z=\frac{1454-1387}{161}=0.416=0.42

P( z = 0.42 ) = 0.6628   (from z-score table)

Thus, P( X ≥ 1454 ) = P( z ≥ 0.42 ) = 1 - 0.6628 =  0.3372

Therefore, Probability that a randomly selected broiler weighs more than 1454 g is 0.3372 or 34% (approx.)

8 0
3 years ago
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